Volume form on Hamiltonian level surface Assume we have a Hamiltonian system on $(\mathbb{R}^{2n},\omega)$ with Hamiltonian $H = H(q,p)$. In a paper I read, it says, without clarification, that the natural Liouville measure $\mu$ obtained by the volume form $\Omega = \omega \wedge \cdots \wedge \omega $ restricts on the energy level surfaces $N = \left\{(q,p) \in \mathbb{R}^{2n}: H(q,p) = c\right\}$ to 
$$\iota(F(q,p))\Omega$$
where F(q,p) is a function that satisfies $\omega(X_{H},F(q,p)) = dH(F(q,p)) = 1$ and $\iota(\cdot)$ is the interior product. A possible choice would be $F(q,p) = \frac{grad H}{\|grad H\|}$.
Why is this so?
Thanks in advance!
 A: Given a hypersurface $S$ in a manifold $M$ with volume form $\omega$ you can construct the volume element on $S$ induced by $\omega$ as follows: Take a field of unit normal vectors $N$ along $S$ and contract $\omega$ by $N$ that is define the differential form of 1 less rank as 
$$\omega_S(\cdot,...)= \omega(N,\cdot,...)$$
The resulting element is called the volume (or sometimes area) element on $S$ induced by $\omega$. 
In this setting we want however a volume form on $S$ which is invariant under the flow. So the usual volume element may not work for us. Now first to construct the volume element on an energy surface, we note that $\frac{\nabla H}{|\nabla H|}$ is unit vector field normal to $S$ (since $S$ is a constant level set of the function $H$). Therefore the volume element on $S$ is given as
$$\omega_S(\cdot,...) = \omega(\frac{\nabla H}{|\nabla H|},\cdot,...)$$
This means that we can write
$$\omega = \omega_s \wedge \frac{dH}{|\nabla H|}=\frac{\omega_S}{|\nabla H|}\wedge dH$$
Now under your flow $\phi$, the forms $\omega$ and $dH$ are invariant that is $\phi^*\omega= \omega$ and $\phi^*dH =dH$. Then
$$\omega= \phi^* \omega = \phi^*\frac{\omega_S}{|\nabla H|} \wedge \phi^*dH$$
So we get that
$$ \frac{\omega_S}{|\nabla H|}\wedge dH =  \phi^*\frac{\omega_S}{|\nabla H|} \wedge dH$$
Now take any collection of basis vectors $v_i$ from the tangent space of $TS$ and contract both sides. Since $dH(v_i)=0$ what we get is that
$$ \frac{\omega_S}{|\nabla H|}(v_1,...,v_k) dH =  \phi^*\frac{\omega_S}{|\nabla H|} (v_1,...,v_k)dH$$
for all vector $v_i$ which implies that 
$$\frac{\omega_S}{|\nabla H|} = \phi^*\frac{\omega_S}{|\nabla H|}$$
so $\frac{\omega_S}{|\nabla H|}$ is invariant.
